# Math (Calculus 121)

posted by
**drwls**
.

The derivative tells you the slope at any point.

df/dx = 8 x^3 - 12 x

When x = 3, the slope is

m = 8*27 - 36 = 180

and the equation of the tangent line is

(y - 228) = 180*(x - 3)

y = 180 x - 540 + 228 = 180 x - 312

wow, that was absolutely correct (i just plugged it into webassign), thanks a lot!!

please just help me to understand the last part though for the test, where did you get the (y-228) = 180*(x-3)?

Find the equation of the line that is tangent to the graph of f(x) = 2x4 - 6x2 + 120 at the point (3, 228).

y=?

The equation for a line with slope m through points x = a and y = b is

(y- b)= m (x - a)

That is where the formula came from.